# Solve $y'+2xy^2 = 1$

Solve $$y'+2xy^2 = 1$$

I've tried to test if the equation is exact or if it can be made exact. Nothing worked and then I understood that it's a non-linear equation. I've tried a lot to think of a substitution so that I can do separation of variables but that also didn't work. It's also not even a Bernoulli's equation.

This might be a simple problem for some but I can't get my head around it. Can I please get some hint? I'm not asking for a full solution so please don't close this question. I can't solve it and I need some help doing it on my own.

• Look at this post math.stackexchange.com/q/3761357/399263 it is a Ricatti ODE reducible to Bessel ODE with probably ugly looking solutions...
– zwim
Commented May 10, 2022 at 20:20
• @Itachi: as zwim mentions, this very simple looking ODE has a ugly solution! See: wolframalpha.com/input?i=y%27+%2B+2+x+y%5E2+%3D+1 . A direction field plot will show why the analytic solution is not simple.
– Moo
Commented May 10, 2022 at 20:22
• Then I think there must be a mistake. This kind of method aren't even talked about in my class. It is indeed an ugly looking solution. Thank you @Moo Commented May 10, 2022 at 20:26
• Why not to try a series solution ? Commented May 11, 2022 at 2:54

This is not a full solution, just some pointers which might help on the way towards a solution.

This is a Riccati equation which according to Wolfram Alpha has solutions in so called Airy and Bairy functions which are defined as solutions to the related family of differential equations.

$$\frac{d^2y}{dy^2} -xy = 0$$

These solutions can not be expressed in elementary functions, but are rather commonly expressed as trigonometric integrals $$Ai(x) = \frac 1 \pi \int_0^\infty \cos\left(\frac{t^3}3 +xt\right)dt$$

Moreover, we can easily see by ocular inspection that there can't be a polynomial solution to our question.

This is because the $$y'$$ shrinks in order, while $$2xy^2$$ grows (as compared to $$y$$).

If we slightly modify the differential equation to $$y' + y^2 = 1$$ This becomes a separable equation: $$\frac{y'}{(1+y)(1-y)} = 1$$ which we can integrate $$\int\frac{dy}{(1+y)(1-y)} = \int 1 dx$$

This gives us a hint that exponential functions shall be involved as the left hand sides gives us logarithms in $$y$$ (after some partial fraction decomposition).

How to handle the $$2x$$ factor, I don't know. Maybe we can do $$\sqrt{2x}$$ and a variable substitution of some kind.

To transform the original nonlinear differential equation into a linear one, we make the substitution $$y=\varphi\frac{u'}{u}$$, where $$\varphi$$ is a function to be determined. Thus, $$y'+2x^2y=1 \implies \varphi'\frac{u'}{u}+\varphi\frac{u''}{u}+(2x\varphi^2-\varphi)\frac{(u')^2}{u^2}=1. \tag{1}$$ Choosing $$\varphi(x)=\frac{1}{2x}$$, we eliminate the nonlinear term $$\frac{(u')^2}{u^2}$$ from $$(1)$$, which then we can rewrite as $$-\frac{1}{2x^2}\frac{u'}{u}+\frac{1}{2x}\frac{u''}{u}=1 \implies xu''-u'-2x^2u=0. \tag{2}$$ Although not obvious, Eq. $$(2)$$ is related to Airy equation, $$z''-tz=0, \tag{3}$$ whose general solution is $$z(t)=c_1\text{Ai}(t)+c_2\text{Bi}(t)$$. Indeed, differentiating $$(3)$$ one obtains $$z'''-tz'-z=0 \implies z'''-tz'-\frac{z''}{t}=0 \implies tz'''-z''-t^2z'=0. \tag{4}$$ One can transform Eq. (2) into Eq. (4) with the substitutions $$u(x)=z'(t)$$, $$t=\sqrt[3]{2}x$$. Therefore, the solution to Eq. $$(2)$$ is $$u(t)=c_1\text{Ai}'(\sqrt[3]{2}x)+c_2\text{Bi}'(\sqrt[3]{2}x). \tag{5}$$ Plugging $$(5)$$ into $$y=\varphi\frac{u'}{u}$$, we finally obtain $$y(x)=\frac{1}{2x}\frac{\sqrt[3]{2}[c_1\text{Ai}''(\sqrt[3]{2}x)+c_2\text{Bi}''(\sqrt[3]{2}x)]}{c_1\text{Ai}'(\sqrt[3]{2}x)+c_2\text{Bi}'(\sqrt[3]{2}x)}=2^{-1/3}\frac{c_1\text{Ai}(\sqrt[3]{2}x)+c_2\text{Bi}(\sqrt[3]{2}x)}{c_1\text{Ai}'(\sqrt[3]{2}x)+c_2\text{Bi}'(\sqrt[3]{2}x)}, \tag{6}$$ where we used $$(3)$$ to simplify the numerator of the last fraction.